Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week14_changing_current_rectangle [2017/11/28 15:34] – [Solution] tallpaul
+++ 184_notes:examples:week14_changing_current_rectangle [2021/07/13 13:26] (current) – [Solution] schram45
@@ Line 1: / Line 1: @@
+[[184_notes:b_flux_t|Return to Changing Magnetic Fields with Time notes]]
 ===== Changing Current Induces Voltage in Rectangular Loop =====
 Suppose you have an increasing current through a long wire,  $I(t) = I_0 \frac{t}{t_0}$ . Next to this wire, there is a rectangular loop of width  $w$  and height  $h$ . The side of the rectangle is aligned parallel to the wire so that the rectangle is a distance  $d$  from the wire, and they are both in the same plane. What is the induced voltage in the rectangle? In what direction is the induced current in the rectangle?
@@ Line 12: / Line 14: @@
 ===Approximations & Assumptions===
-  * The long wire is infinitely long and thin and straight.
+  * The long wire is infinitely long and thin and straight: With these assumptions the magnetic field produced by the wire only depends on the radial distance away from the wire and the current in it. This also allows us to use a simplified magnetic field equation from the notes.
-  * There are no external contributions to the B-field.
+  * There are no external contributions to the B-field: We are not told about any other external moving charges or currents that could also produce a magnetic field that would effect the flux through our loop, so we will assume the only contribution is from the long wire.
 ===Representations===
@@ Line 21: / Line 23: @@
   * We represent the situation with the following visual. We arbitrarily choose a direction for the current.
-{{ 184_notes:14_wire_rectangle.png?300 |Wire and Rectangle}}
+[{{ 184_notes:14_wire_rectangle.png?500 |Wire and Rectangle}}]
 ====Solution====
-In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the right hand rule to determine the the magnetic field from the wire is into the page ( $-\hat{z}$ ) near the rectangle. For convenience, we will also define for the area vector to be into the page. Since they both point in the same direction, the dot product simplifies:
+In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the [[184_notes:rhr|right hand rule]] to determine that the magnetic field from the wire is into the page ( $-\hat{z}$ ) near the rectangle (point your fingers in the direction of the current and curl them toward the rectangle - your thumb should point into the page). For convenience, we will also define for the area vector to be into the page. Since they both point in the same direction, the dot product simplifies:
  $\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$
-Usually, we would pull the  $B$  out of the integral, but we cannot do that in this case! That is because  $B$  varies for different points in the rectangle's area. In order to see this a little more clearly, we will break the integral down a little and also insert the expression for the magnetic field from a long wire:
+Usually, we would pull the  $B$  out of the integral, but **we cannot do that in this case**! That is because  $B$  varies for different points in the rectangle's area - the magnetic field would be stronger close to the wire and get weaker as you move to the parts of the rectangle that are farther away. In order to see this a little more clearly, we will break the integral down a little and also insert the expression for the magnetic field from a long wire:
  $\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$
@@ Line 33: / Line 35: @@
 In the first equality, we just changed the integral from over the "rectangle" to be over its two dimensions, "left to right", and "top to bottom". We also changed  $\text{d}A$  to be in terms of its two dimensions,  $\text{d}A = \text{d}x\text{d}y$ . A visual is shown below for clarity.
-{{ 184_notes:14_da_dx_dy.png?600 |Picture of dA Breakdown}}
+[{{ 184_notes:14_da_dx_dy.png?500 |Picture of dA Breakdown}}]
 At this point, our integral is set up enough that we can crunch through the analysis. We can pull out constants to the front and calculate to the end:
 \begin{align*}
-\int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y &= \frac{\mu_0 I}{2 \pi} \int_{d}^{d+w} \frac{\text{d}x}{x} \int_{0}^{y} \text{d}y \\
+\int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y &= \frac{\mu_0 I}{2 \pi} \int_{d}^{d+w} \frac{\text{d}x}{x} \int_{0}^{h} \text{d}y \\
-&= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{y} \\
+&= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{h} \\
 &= \frac{\mu_0 I}{2 \pi} \left(\log(d+w) - \log(d)\right) \left( h-0 \right) \\
 &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right)
 \end{align*}
-At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant in time, //except// for to current,  $I$ . Here is the induced voltage:
+At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant with respect to time, //except for the current  $I$ //. The induced voltage can then be found by taking the time derivative (to find the changing magnetic flux through the rectangular loop):
- $V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$
-We wish to find the magnetic field in the plane we've shown in the representations. Due to the circular symmetry of the problem, we choose a circular loop in which to situate our integral  $\int \vec{B}\bullet\text{d}\vec{l}$ . We also choose for the loop to be the perimeter of a flat surface, so that the entire thing lies in the plane of interest, and there is no enclosed current (so  $I_{enc} = 0$ ). We show the drawn loop below, split into two cases on the radius of the loop.
-{{ 184_notes:14_capacitor_loops.png?600 |Circular Loops}}
-Below, we also draw the direction of the magnetic field along the loops. We know the magnetic field is directed along our circular loop -- if it pointed in or out a little bit, we may be able to conceive of the closed surface with magnetic flux through it, which would imply the existence of a magnetic monopole. This cannot be the case! We also know that the field is directed counterclockwise, due to the increasing electric field into the page. This comes from an extension of Lenz's Law, upon which discussion is not needed for this course.
-{{ 184_notes:14_capacitor_b_field_loops.png?600 |Circular Loops, with B-field shown}}
-We are pretty well set up to simplify our calculation of the integral in the representations, since the B-field is parallel to the loop's perimeter. Below, we show the integral calculation, where the magnetic field at a radius  $r$  is displayed as $B(r)$.
- $\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$
-Next, we need to find the changing electric flux in our loop. Since our loop was described with a flat surface, and the electric field is directed parallel to the area-vector of the loop, we can write electric flux as  $\Phi_E = \vec{E} \bullet \vec{A} = EA$ . This formula will need to be split up for parts of the surface inside the plates versus outside, since the electric field is different.
- $\Phi_\text{E, in} = EA = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{loop}} = \frac{Q}{\epsilon_0\pi R^2}\pi r^2 = \frac{Qr^2}{\epsilon_0 R^2}$
- $\Phi_\text{E, out} = EA = E_\text{in}A_\text{in} + E_\text{out}A_\text{out} = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{plate}} + 0 = \frac{Q}{\epsilon_0\pi R^2}\pi R^2 = \frac{Q}{\epsilon_0}$
-Now, if we wish the find the change in flux, we will take a time derivative. Notice that all the terms in the flux expressions above are constant, except for  $Q$ , which is changing with time as dictated by  $I$ .
- $\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$
- $\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$
-We can now connect the pieces together (remember,  $I_{enc}=0$ , so we omit it below). We can write:
- $2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$
- $2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$
-We are ready to write out the magnetic field.
-\[
+$$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$
-B(r) = \begin{cases}
-          \frac{\mu_0 I r}{2\pi R^2} &&& r<R \\
-          \frac{\mu_0 I}{2\pi r} &&& r>R
-       \end{cases}
-\]
-Notice, the distance between the plates has no effect on the magnetic field calculation. Also, the amount of the charge on the plates at a given time does not matter -- we only care about how fast the charge is changing (the current!). Also, it is interesting that outside the plates, the magnetic field is the same as it would be for a long wire. This would be just as if the capacitor were not there, and the wire were connected. Below, we show a graph of the magnetic field strength as a function of the distance from the center of the capacitor.
+Notice that the induced voltage is negative. This means that the induced current produces a magnetic fields whose corresponding flux is negative flux. Since the area-vector was defined as into the page ( $-\hat{z}$ ), this means that the magnetic field produced by the induced current should be out of the page ( $+\hat{z}$ ). By the right hand rule, this means the induced currents is directed //counterclockwise//. See below for a visual.
-{{ 184_notes:14_capacitor_b_field_graph.png?400 |B-Field Strength, Graphed}}
+[{{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}}]
-We have enough information to find the maximum B-field, which is at the edge of the plates:
+The induced voltage in this problem was a constant, which means the flux must be changing linearly with time. Thats the only way the derivative can be a constant. If we look at the equation for flux through a loop there are three ways it can change. The first is if the magnetic field changes, the second is if the cross product changes, and the third is if the area changes with time. None of the sides of our loop are moving, so the area is not changing with time. The relationship of our magnetic field and area vector are also not changing as our loop is not rotating and our magnetic field is always in the same direction. This leaves the magnetic field, which is changing in this example. The magnetic field in this problem changes just like the current in the long wire. Since the current in the long wire changes linearly with time, our flux must change linearly with time as well. This means our induced voltage should be a constant as our solution shows.
- $B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$